Problem: $~ f(x)=$ $\frac{1}{1+2x}$ We know that $f(x)=1-2x+{{(2x)}^{2}}-{{(2x)}^{3}}+...+{{(-2x)}^{n}}+...$ for $x\in \left(-\dfrac12,\dfrac12\right)$. Using this fact, find the power series for $g(x)=-\dfrac{2}{(1+2x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-2+8x-24{{x}^{2}}+...$ (Choice B) B $2+8x+24{{x}^{2}}+...$ (Choice C) C $-2+4x-6{{x}^{2}}+...$ (Choice D) D $1+2x+3{{x}^{2}}+...$ (Choice E) E $2-8x+24{{x}^{2}}+...$
Answer: First note that $~g(x)=f\,^\prime(x)\,$. Hence, to find the series represented by $~g(x)\,$, differentiate the series for $~f(x)~$ term-by-term. With $f(x)=1-2x+4x^2-8x^3+16x^4-...\,$, we get $ g(x)=f\,^\prime (x)=-2+8x-24{{x}^{2}}+64x^3-...$